$\begin{array}{l}y = \sqrt{\dfrac{1}{sin^2x - 1}}\\ \text{y xác định}\\ \Leftrightarrow \dfrac{1}{sin^2x - 1} \geq 0\\ \Leftrightarrow sin^2x - 1 < 0\\\Leftrightarrow -1 < sinx < 1\\\Leftrightarrow x \in \left(-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right) + k2\pi \,\,\,(k \in \Bbb Z)\end{array}$